Unable to generate a key of random letters which are only used/found once in the entire key

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I am attempting to generate a randomized key of letters which are only used once in the overall key. I currently have an array "regKey" which is storing the letters A-Z in its normal order. I would like to create a new array, "newKey" where the order of the letters are completely random but each and every letter is used when creating this new array. There should be no duplicates of any letter in the new array.

So far I've been able to generate a random key but often there are duplicates of certain letters. Here is my code below for reference.

public void keyGen() {
             char [] regKey = {'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'}; 
             char [] newKey = new char [26];

             int tempNum;
             int totalChoice = 26;

             Random rand = new Random();

             for(int i = 0; i<26; i++) {
                 tempNum = rand.nextInt(totalChoice);

                 newKey[i] = regKey[tempNum];

                 System.out.print(newKey[i]);
             }

             String keyString = new String (newKey);

             label_key.setText(keyString);
        }

Assuming you definitely need an Array in the first place, the following code does create the output you desire:

char [] regKey = {'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'}; 
char [] newKey = new char [26];
String[] array = new String(regKey).split("", 0);
ArrayList<String> yourNewArrayList = new ArrayList<String>();
Collections.addAll(yourNewArrayList, array);
Collections.shuffle(yourNewArrayList);
for (int i = 0; i < newKey.length; i++) {
    newKey[i] = yourNewArrayList.remove(0).toCharArray()[0];
}

Generating a random & unique 8 character string using MySQL , I thought of the generate&check loop approach again, but I'm not limiting this question to that just in case there's a more efficient one. I've been able to generate� The operation you are looking for is called a shuffle or a permutation. It is not sufficient to call a random-letter function 26 times, since, as you see, you can generate duplicates. Instead, start with the string "ABCDEFGHIJKLMNOPQRSTUVWXYZ" and perform a shuffle operation.

What are you trying to do is called "shuffle". You can use Collections.shuffle to "randomize" an array.

List<Character> regKeyAsList = Arrays.asList(regKey);
Collections.shuffle(regKeyAsList);
char[] newKey = regKeyAsList.toArray();

Saving a file in Chrome my keyboard types random letters and , Just encountered the same issue in Chrome and Brave --NOT in Edge. And yes-- disabling Alert key stroke encryption fixed it. Go to advanced� To generate a random password you can use pwgen: pwgen generates random, meaningless but pronounceable passwords. These passwords contain either only lowercase letters, or upper and lower case mixed, or digits thrown in. Uppercase letters and digits are placed in a way that eases remembering their position when memorizing only the word.

Either with Collections.shuffle. Here using Unicode code points.

List<Integer> alphabet = IntStream.rangeClosed('A', 'Z').boxed()
                             .collect(Collection.toList());
// Or enumerate all:
// List<Integer> alphabet = IntStream.of('A', 'B', 'C', ..., 'X', 'Y', 'Z').boxed()
//                              .collect(Collection.toList());
Collections.shuffle(alphabet);
String newKey = new String(alphabet.stream()
        .mapToInt(Integer::intValue)
        .toArray(), 0, alphabet.size());

Or:

         for (int i = 0; i < 26 - 1; i++) {
             int pickedI = i + rand.nextInt(26 - i);

             // Swap i and pickedI:
             int old = newKey[i];
             newKey[i] = regKey[pickedI];
             newKey[pickedI] = old;

             System.out.print(newKey[i]);
         }

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The one-time pad (video) | Cryptography, It can only be used if you can meet the other person and transfer the pad to Another problem Duration: 2:56 Posted: Apr 21, 2012 The company I used to work for example used a hardware random number generator which made use of the fundamentally random nature of quantum optics as a source of true randomness. It was USB-interfaced an allowed random data streams with a speed up to 16 Mbps. We used that hardware as an ENGINE for OpenSSL to create our ECC and RSA keys.

rand - Manual, Note: As of PHP 7.1.0, rand() uses the same random number generator as mt_rand(). means you only need three consecutive values to be able to predict its entire //and we assure no number is used more than once (technically reiterating prev line). //"0" because we use this to FIND ARRAY KEYS which has a 0 value Instead, they are pseudo-random. The return values should only be used in case true randomness is not that important, such as in the classic number-guessing game. In case you need a random value to be used in cryptography such as a cryptographic key in symmetric and asymmetric encryption then System.Random is not an acceptable option.

Comments
  • Does this answer your question? Random shuffling of an array
  • Use a List to store the characters and remove the character you used so it won't be available when generating the next one.
  • @JoakimDanielson I just provided the whole code for achieving the end result, including the transformation.
  • How can you have a List of char, char is a primitive type?
  • Is it required to create a list or can this be done by simply using Collections.shuffle(name_of_array); ?
  • Why are people up-voting an answer that doesn't even compile? Please fix your answer or remove it.
  • I am encountering the same issue, char cannot be resolved as the dimensions in the list
  • @BipoN Check out my answer below, I provided a working solution tackling the issue of List with char type
  • Neither of your answers compiles and if I fix the second one it produces a result with duplicates.
  • @JoakimDanielson Appreciate your feedback that I failed here. About the second '(easiest) example: it does not print the last element (i < 26-1) as that needs no randomizing. But if newKey is filled by regKey with unique entries, swapping two elements will not cause duplicates.